Question

The Standard says that std::tuple has the following member functions

constexpr tuple();
explicit tuple(const Types&...);

Can someone please explain what is supposed to happen for std::tuple<>?

Answer 1

I believe this is a minor error in the standard. Clearly, when the Types parameter pack is empty, the two constructor calls are equivalent and cannot be overloaded (see C++11 section 13). (Further note that the constructor using Types is not a member template either –if it was, then it would be a legal overload.).

In other words, this code will not compile:

template <typename... Types>
struct Test
{
  constexpr Test() {}
  explicit Test(Types const&...) { /* etc. */ }
};

int main()
{
  Test<> a;
  Test<int> b;
}

e.g., a g++ v4.8 snapshot outputs:

tt.cxx: In instantiation of ‘struct Test<>’:
tt.cxx:10:10:   required from here
tt.cxx:5:12: error: ‘Test<Types>::Test(const Types& ...) [with Types = {}]’ cannot be overloaded
   explicit Test(Types const&...) { /* etc. */ }
            ^
tt.cxx:4:13: error: with ‘constexpr Test<Types>::Test() [with Types = {}]’
   constexpr Test() {}
             ^

This can be fixed by using partial specialization:

template <typename... Types>
struct Test
{
  constexpr Test() {} // default construct all elements
  explicit Test(Types const&...) { /* etc. */ }
  // and all other member definitions
};

template <>
struct Test<>
{
  constexpr Test() {}
  // and any other member definitions that make sense with no types
};

int main()
{
  Test<> a;
  Test<int> b;
}

which will compile correctly.

It appears the standard wanted a constexpr default constructor was so that std::tuple<> var; could be written instead of writing std::tuple<> var(); or std::tuple<> var{}; because of the use of explicit with the other constructor. Unfortunately, its definition of std::tuple does not work for tuples of size zero. The standard does permit such in section 20.4.2.7 (relational operators) though, “For any two zero-length tuples, […]”. Oops! 🙂